Let \(\overrightarrow{\mathrm{a}}=2 \hat{i}-\hat{j}+3 \hat{k}, \overrightarrow{\mathrm{~b}}=3 \hat{i}-5…
- 462
- 77
- 154
- 308
Solution
& \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{~b}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}} \\ & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=0 \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow \overrightarrow{\mathrm{c}}=\lambda(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \\ & \overrightarrow{\mathrm{c}}=\lambda(5 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \ldots . .(1) \\ & |\overrightarrow{\mathrm{c}}|^2=\lambda^2(25+36+16) \\ & |\overrightarrow{\mathrm{c}}|^2=77 \lambda^2 \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}) \cdot(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=168 \\ & \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{~b}}+|\overrightarrow{\mathrm{c}}|^2=168
\end{aligned}$
$14+\vec{c} \cdot(\vec{a}+\vec{b})+77 \lambda^2=168$
using equation (1)
$\begin{aligned}
& \lambda|5 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}|^2+77 \lambda^2=154 \\ & 77 \lambda+77 \lambda^2-154=0 \\ & \lambda^2+\lambda-2=0 \\ & \lambda=-2,1
\end{aligned}$
$\therefore$ Maximum value of $|\overrightarrow{\mathrm{c}}|^2$ occurs when $\lambda=-2$
$\begin{aligned}
& |\overrightarrow{\mathrm{c}}|^2=77 \lambda^2 \\ & =77 \times 4 \\ & =308
\end{aligned}$ ,
Asked in: JEE Main 2025 (29 Jan Shift 1)